Suppose I have a random number (eg 625 467 921). I use it once to encrypt a message OTP style, and when I want to use them again, I reverse them (eg to 129 764 526) and use to encrypt the other message. What is the weakness of this scheme?

Again, suppose I have some random numbers (again, eg 625 467 921). When I want to use them again, I add a static value to each number (eg adding $4$ modulo $10$, the numbers will be 069 801 365) and use these new values for the second message. For the 3rd message, I add to the numbers $6$ or $7$ etc. I keep using the same base numbers for each message, and all I need to encrypt is the number to add to the last used numbers.
Will these numbers be random? What is the weakness in the scheme?

What do you assume about the attacker? Does the attacker know the process, just not the initial random number? This would be the most natural assumption if you follow kerchoffs' principle. If you don't follow that principle, you are building on unstable grounds.
–
mikeazo♦Dec 17 '12 at 12:22

1 Answer
1

A standard assumption in cryptography is that an attacker knows everything about your cryptosystem, apart from the key. This is known as Kerckhoffs's principle. Taking a look at your construction under this assumption, shows that it's extremely weak.

is unkeyed and can be trivially reversed by an attacker. So you end up with something that amounts to a two-time-pad, and that's not very secure.

The second one uses a keyed transformation, essentially a caesar cipher. Apart from this transform you're once again reusing a one-time-pad. Since the caesar cipher has a very small key, your security gain over plain one-time-pad reuse is just as small. Still extremely weak.